The same force applied anywhere on a rigid object causes the same translational accelleration.
The difference is that forces not applied in the direction of the center of mass will also cause some rotational accelleration.
Remember that F and A in F=mA are vectors. You can therefore treat the vectors as components in any orthagonal system you like. One way to break up F is in the direction towards the center of mass and in the plane perpendicular to that. Only the part in the plane perpendicular to the direction to the center of mass will cause roational accelleration. The entire F causes translational accelleration.
Another way to look at the applied force is to break it down into the force applied to the mass that only causes translational accelleration, and a torque that only causes rotational accelleration.
The first is simply the force divided by the mass. For the purposes of finding the positional change of the center of mass, it makes no difference where the force is applied. The center of mass will move exactly the same way in your cases 1 and 2.
Torque is the cross product of the vector from the center of mass to where the force is applied, and the force vector. That torque causes rotational accelleration, but has no effect on the position of the center of mass. In your case 1, the torque is 0 since the vector from the center of mass to the force application point is zero. If you want to say that the rod has some thickness and that therefore the force is being applied one rod radius left of the center of mass, then the torque is still zero. That is because the force vector and the vector from the center of mass to the application point are parallel, so their cross product is zero. It should also make intuitive sense that case 1 doesn't cause the rod to rotate, just accellerate to the right.
In case 2, the cross product is clearly non-zero. Let's say the rod is 1 m long and the force is 3 N to the right. The vector from the center of mass to the force application point is therefore 1/2 m up. This vector cross the force vector is 1.5 Nm into the plane of your diagram, which will cause a clockwise accelleration around the center of mass. Since this is only a torque, it has no effect on the position of the center of mass.